I couldn't find this very problem anywhere, albeit looking for an hour now...
Simple question: What is $\int \mathrm{d}x f(x) \delta(\mathrm{e}^x)$?
$f(x)$ is any suitable function.
Mathematica spits out zero, but I don't understand why, since $\mathrm{e}^x$ has no zero and hence the usual formula for functions as delta arguments $\delta(g(x)) = \sum_i \frac{\delta(x-x_i)}{|g'(x_i)|}$ [where $x_i$ are zeros of $g(x)$] doesn't work.
I tried to substitute the argument of the delta $\mathrm{e}^x = y$, tried to add an infinitesimal $\epsilon$ in the argument, to get $\delta(\mathrm{e}^x-\epsilon)$ or tried to write the delta as a delta sequence $\delta(x) = \lim_{\epsilon \rightarrow 0} \delta_\epsilon(x)$ and to take the limit $\epsilon \rightarrow 0$ at the end of the calculation.
Nothing worked for me, to show, that the above integral is zero.
You need to first make up your mind what the expression means. A distribution is a functional on $C^\infty_c(\Bbb{R})$ and a tempered distribution is a functional on $\Bbb{S}(\Bbb{R})$ (the Schwartz space).
If you make a change of variable $x=\ln y$ then $$\int_{-\infty}^{+\infty} \text{d}x f(x) \delta(e^x) = \int_0^{+\infty} \text{d}y \frac{f(\ln y)}{y} \delta(y)$$ and presumably the intent is that this last integral is $$\lim_{y\searrow 0} \frac{f(\ln y)}{y} = \lim_{x\rightarrow -\infty} e^{-x} f(x)$$
Now, if $f\in C^\infty_c(\Bbb{R})$ then this limit is zero, because there's an $N\in\Bbb{R}$ such that $f(x) = 0$ for $x<N$.
However, if $f\in\Bbb{S}(\Bbb{R})$, this limit may not exist - because Schwartz functions f are guaranteed to have $|x|^k f(x)$ bounded for all $k>0$, but $e^{-x}$ does not fit within such a bound as $x\rightarrow -\infty$. So the expression is not well-defined for Schwartz functions and is not a tempered distribution. Informally it's because the growth of $\delta(e^x)$ is faster than any polynomial as $x\rightarrow -\infty$.